EXTRA CREDIT PROJECTS
MICHAEL A. TYCHONIEVICH
The following extra credit projects are to be turned in by the end of the day of the last day
before finals start. Each successfully completed extra credit project is worth 5 points, added
to your midterm grade. Some partial credit may be given for incomplete solutions, but it
will not be much. You are encouraged to attend office hours to discuss your solutions with
me before the due date so that you can make corrections. Each of the numbered problems
is considered part of the project. These projects require you to have far-reaching knowledge
of the topics covered in this class and as such each assignment may require a good deal of
time to complete.
Note. As this assignment sheet is a work in progress, it may contain errors; please contact
me if you think that you have found any.
1. Propositional Logic
In this project you will prove that any truth table is the truth table of some propositional
form built up from propositional variables using only the connectives ∧, ∨, and ¬. To do this,
we will need to go into the theory of propositional logic at least far enough to give rigorous
definitions of propositional forms and truth tables. For the problem, we fix countably many
distinct propositional variables A0 , A1 , A2 , . . . and for each n ∈ N we define the set An of the
first n atoms by An = {A0 , A1 , . . . An−1 }.
We define the set Pn of propositional forms involving the first n propositional variables
An recursively as follows. We say that an object P ∈ Pn if and only if one of the following
holds:
• P is one of the propositional variables A0 , A1 , . . . An−1 .
• There is a propositional form Q ∈ Pn such that P is ¬Q.
• There are propositional forms Q ∈ Pn and R ∈ Pn such that P is Q ∨ R.
• There are propositional forms Q ∈ Pn and R ∈ Pn such that P is Q ∧ R.
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For each n ∈ N, we say that a truth assignment is a function v : Pn → {T, F } such that
for any propositional form P ∈ Pn ,
• If there is a propositional form Q ∈ Pn such that P is ¬P , then v(P ) = T if and only
if v(Q) = F .
• If there are propositional forms Q ∈ Pn and R ∈ Pn such that P is Q ∨ R, then
v(P ) = T if and only if either v(Q) = T or v(R) = T .
• If there are propositional forms Q ∈ Pn and R ∈ Pn such that P is Q ∧ R, then
v(P ) = T if and only if both v(Q) = T and v(R) = T .
1.1. Show that a truth assignment is completely determined by what it does to propositional
variables. That is, show that if v and u are truth assignments such that v(Ak ) = u(Ak ) for
every k ∈ {0, 1, . . . , n − 1}, then v = u.
Now let Vn be the set of all truth assignments on Pn . Let Fn be the set of all functions
from Vn to {T, F }. We call Fn the set of all truth tables. Each truth assignment corresponds
to a specific row on a truth table, and a function in Fn corresponds to a column in which each
row contains either a T or an F . For each P ∈ Pn , we define the function fP : Vn → {T, F }
by fP (v) = v(P ), and let En be the set of all such fP . We call the set En the set of all truth
tables that come from propositional forms. Each function in En corresponds to a column
of a truth table that is calculated from a specific propositional form, and the truth assignments to the propositional variables are listed in the far left columns as usual for truth tables.
1.2. Verify that each truth table that comes from a propositional form is an official truth
table. That is, show that for each n ∈ N that En ⊆ Fn .
The set En here allows us to think about propositional forms and their relationship with
truth assignments in a different way. Rather than thinking of the propositional form as the
object and the truth value assignment as the function, we can think of the proposition as
the function and the truth value assignment as the object. Again, when one draws a table
of values for this function, it has the form of a truth table as we discussed in the first weeks
of class. A question arises here: is it possible that there are functions in Fn that don’t
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arise from propositional forms? In other words, are there truth tables that don’t come from
propositional forms? In what follows, you will prove that in fact En = Fn , which means that
all possible truth tables not only come from propositional forms, but that we need at most
the three connectives we started with to get all possible truth tables.
1.3. There is a propositional form P ∈ Pn such that for all truth assignments v ∈ Vn ,
v(P ) = F .
1.4. Let v ∈ Vn be a truth value assignment such that there is a m ∈ {0, 1, . . . , n − 1}
such that for all k ∈ {0, 1, . . . , n − 1}, v(Ak ) = T if and only if k = m. Then there is a
propositional form P ∈ Pn such that for all truth assignments u ∈ Pn , u(P ) = T if and only
if u = v. [Hint: Do a proof by induction on n. For the induction step, take an appropriate
propositional form Q ∈ Pn−1 and define P to be either Q ∧ An or Q ∧ (¬An ).]
1.5. Let V ⊆ Vn be a set of truth value assignments. Then there is a propositional form
P ∈ Pn such that for all truth assignments v ∈ Vn , v(P ) = T if and only if v ∈ V .
[Hint: Prove this by induction on the number of elements of V , forming your base case
from 1.3 and 1.4.]
1.6. For each truth table f ∈ Fn , there is a propositional form P ∈ Pn such that f = fP .
Thus En = Fn . [Hint: Let V be the set of all truth value assignments v such that f (v) = T ,
and then apply 1.5 to the set V . Use 1.2 to conclude that En = Fn .]
We now know that En = Fn for any natural number n. A natural question to ask at this
point is this: what about infinite truth tables? It turns out that not all infinite truth tables
are given by finite propositional forms. To prove this we need a few more definitions. Let
P = ∪∞
n=0 Pn and let V be the set of all truth assignments on P. Let F be the set of all
functions from V to {T, F }. For each P ∈ P, define fP ∈ F by fP (v) = v(P ).
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1.7. Let f ∈ F be the infinite truth table defined by
f (v) =


T, if v(An ) = T for all n ∈ N,

F, otherwise,
for each v ∈ V. (i.e. the first row of the truth table f has a T in it, and all other rows
have F .) Show that there is no propositional form P ∈ P such that f = fP . [Hint: Suppose
toward a contradiction that P is a propositional form such that f = fP . Show by the definition of P that P ∈ Pn for some natural number n. But P is also an element of Pn+1 . Use
this to show that for any truth assignment, changing only the truth value of An+1 cannot
affect the truth value of P . Show that this fact contradicts f = fP because there is a truth
assignment such that changing its value at any propositional variable changes the value of f .]
2. The Ackermann Hierarchy
In this project you will analyze some of the behavior of the so-called Ackermann hierarchy
of functions. There is a useful Wikipedia page on this function which you may refer to,
though all proofs in this assignment must be your own. The mth function in the Ackermann
hierarchy is defined recursively as follows:



n + 1,
if m = 0



Am (n) = Am−1 (1),
if m > 0 and n = 0




A
m−1 (Am (n − 1)), if m > 0 and n > 0.
Notice that this definition requires the use of double induction to make sense of it. In fact
2.1. Show that the function Am : N → N is defined for each n ∈ N. [Hint: do a double
complete induction on n and m.]
To understand how the Ackermann functions work, it’s a good idea to work formulas for
the first few functions explicitly.
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2.2. Prove that for each n ∈ N:
(1) A0 (n) = n + 1
(2) A1 (n) = n + 2
(3) A2 (n) = 2n + 3
(4) A3 (n) = 2n+3 − 3
Beyond m = 3, the functions Am increase so fast that they require new notation to make
work with. We won’t go into that notation (you can check out the Wikipedia page on
Knuth’s up-arrow notation if you are curious), but we will go over some basic properties of
the Ackermann hierarchy.
2.3. Prove that each function Am is increasing, and that for each k, m, n ∈ N, if k ≤ m then
Ak (n) ≤ Am (n).
2.4. Define the function A : N → N by A(n) = An (n). Show that A cannot be one of the
functions in the Ackermann hierarchy by showing that for each m ∈ N, there is an N ∈ N
such that A(N ) > Am (N ).
2.5. Show that for any m ∈ N, there is an N ∈ N such that for all n > N , we have that
A(n) > Am (n).
The function A is an extremely fast growing function, and an extremely hard function to
compute. This is especially true if you use the recursion given above to compute it, because
the only operations used in the recursion are the addition and subtraction of the number 1.
This leads us to this final result that may be of interest to anyone who wants to work with
hard-to-compute functions:
2.6. Prove that for each n ∈ N that the computation of A(n) requires more than A(n)
allowed operations, where an allowed operation is either the addition or subtraction of the
number 1. [Hint: since there is no explicit recursion given for A, it may be helpful to look
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at the entire hierarchy of functions to prove this result.]
Now that you have worked with this double recursion, you can apply some of the ideas
used in this problem to recursively define addition, multiplication, and exponentiation of
integers in terms of adding and subtracting the number 1.
2.7. For each part of this problem, give the requested definition under the requested conditions. You need not prove that the definition you give is correct.
(1) Give a recursive definition for n + m using only iteration and the addition and subtraction of the number 1.
(2) Give a recursive definition for n · m using only using only iteration, addition, and
subtraction.
(3) Give a recursive definition for nm using only using only iteration, addition, subtraction, and multiplication.
3. The Axiom of Choice
In this project you will prove that the axiom of choice is equivalent to the statement that
all surjective functions have right inverses. Since we already did part of this proof in class, it
only remains to show that the existence of right inverses for surjective functions implies that
the axiom of choice is true. Given an indexed family of sets, we are going to try to set up a
function so that any right inverse for it is either the required choice function, or something
very much like it. The complication here is that we can’t use the indexing function directly,
so we have to make up a new indexing function to do the job by forcing it to carry around
a bunch of extra information in the index. To do this, we’ll make the set of indices a lot
larger without changing what sets are indexed. At the end of the proof, we’ll apply the right
inverse property of surjective functions to this augmented indexing function to carve away
most of that extra stuff, leaving behind just enough of it to establish a choice function for
the family of sets we started with.
3.1. Let B be a set and let hBα iα∈A be an indexed family of nonempty subsets of B with
index set A. Let B = {Bα : α ∈ A} be the range of this indexed family of sets. Let
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M ⊆ A × B be given by M = {(α, β) : α ∈ A and β ∈ Bα }. For each (α, β) ∈ M , let
B(α,β) = Bα . Show that hBm im∈M is an indexed family of sets with index set M and that its
range is B.
3.2. Since the indexing function is surjective onto its range, it must have a right inverse
g : B → M . Let f : A → B be the function that sends each α ∈ A to the β ∈ B such that β
is the second coordinate of g(Bα ). Prove that the operation just described actually defines
f as a function, and show that f is in fact a choice function for hBα iα∈A .
It turns out that there are many useful equivalent statements of the axiom of choice. Some
versions, such as Tychonoff’s theorem and Zorn’s lemma, are in common use in many areas
of mathematics. As the theoretical build-up needed to make sense of these versions of the
axiom of choice is outside the scope of this class, we’ll finish with a version that only requires
notions that we have previously discussed. Note that this version does not make explicit
mention of functions, but it does rely on the fact that the axiom of choice is always true
whenever every set in an indexed family is finite.
3.3. Show that the following statement is equivalent to the axiom of choice: For any
nonempty set M and any nonempty pairwise disjoint collection M of nonempty subsets
of M , there is a set X ⊆ M such that for each set A ∈ M the intersection A ∩ X has exactly
one element. [Hint: To go from the axiom of choice to this result, use M as your index set
and let X be the image of your choice function. For the converse, you’ll have to somehow
turn your indexed family of sets into a pairwise disjoint collection of sets. It may be useful
to think about the index set M defined in 3.1.]
Department of Mathematics, The Ohio State University, 231 West 18th Avenue, Columbus, Ohio 43210, USA
E-mail address: tycho@math.ohio-state.edu
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